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The boundary line shown is $2x+3y=6$ . Write the inequality shown by the graph.
The line $2x+3y=6$ is the boundary line. On one side of the line are the points with $2x+3y>6$ and on the other side of the line are the points with $2x+3y<6$ .
Let’s test the point $\left(0,0\right)$ and see which inequality describes its side of the boundary line.
At $\left(0,0\right)$ , which inequality is true:
So the side with $\left(0,0\right)$ is the side where $2x+3y<6$ .
(You may want to pick a point on the other side of the boundary line and check that $2x+3y>6$ .)
Since the boundary line is graphed as a dashed line, the inequality does not include an equal sign.
The graph shows the solution to the inequality $2x+3y<6$ .
Write the inequality shown by the shaded region in the graph with the boundary line $x-4y=8$ .
$x-4y\le 8$
Write the inequality shown by the shaded region in the graph with the boundary line $3x-y=6$ .
$3x-y\le 6$
Now, we’re ready to put all this together to graph linear inequalities.
Graph the linear inequality $y\ge \frac{3}{4}x-2$ .
The steps we take to graph a linear inequality are summarized here.
Graph the linear inequality $x-2y<5$ .
First we graph the boundary line $x-2y=5$ . The inequality is $<$ so we draw a dashed line.
Then we test a point. We’ll use $\left(0,0\right)$ again because it is easy to evaluate and it is not on the boundary line.
Is
$\left(0,0\right)$ a solution of
$x-2y<5$ ?
The point
$\left(0,0\right)$ is a solution of
$x-2y<5$ , so we shade in that side of the boundary line.
What if the boundary line goes through the origin? Then we won’t be able to use $\left(0,0\right)$ as a test point. No problem—we’ll just choose some other point that is not on the boundary line.
Graph the linear inequality $y\le \mathrm{-4}x$ .
First we graph the boundary line
$y=\mathrm{-4}x$ . It is in slope–intercept form, with
$m=\mathrm{-4}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}b=0$ . The inequality is
$\le $ so we draw a solid line.
Now, we need a test point. We can see that the point $\left(1,0\right)$ is not on the boundary line.
Is
$\left(1,0\right)$ a solution of
$y\le \mathrm{-4}x$ ?
The point $\left(1,0\right)$ is not a solution to $y\le \mathrm{-4}x$ , so we shade in the opposite side of the boundary line. See [link] .
Some linear inequalities have only one variable. They may have an x but no y , or a y but no x . In these cases, the boundary line will be either a vertical or a horizontal line. Do you remember?
Graph the linear inequality $y>3$ .
First we graph the boundary line $y=3$ . It is a horizontal line. The inequality is>so we draw a dashed line.
We test the point $\left(0,0\right)$ .
$\left(0,0\right)$ is not a solution to $y>3$ .
So we shade the side that does not include (0, 0).
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